Question: Let $g(x)=x^5-3x^2+4x$. Find $g'(-2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $88$ (Choice B) B $-52$ (Choice C) C $96$ (Choice D) D $26$
Let's first find the expression for $g'(x)$ and then evaluate it at $x=-2$. According to the sum rule, the derivative of $x^5-3x^2+4x$ is the sum of the derivatives of $x^5$, $-3x^2$, and $4x$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\dfrac{d}{dx}(x^5)=5x^4$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(x^5-3x^2+4x) \\\\ &=\dfrac{d}{dx}(x^5)-3\dfrac{d}{dx}(x^2)+4\dfrac{d}{dx}(x)&&\gray{\text{Basic differentiation rules}} \\\\ &=5x^4-3\cdot2x+4\cdot 1x^0&&\gray{\text{The power rule}} \\\\ &=5x^4-6x+4 \end{aligned}$ So we found that $g'(x)=5x^4-6x+4$. Plugging in $x=-2$ and evaluating using the calculator, we find that $g'(-2)=96$. In conclusion, $g'(-2)=96$.